P38_046 - 46 The q in the denominator is to be interpreted as |q |(so that the orbital radius r is a positive number We interpret the given 10.0

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46. The q in the denominator is to be interpreted as | q | (so that the orbital radius r is a positive number). We interpret the given 10 . 0 MeV to be the kinetic energy of the electron. In order to make use of the mc 2 value for the electron given in Table 38-3 (511 keV = 0 . 511 MeV) we write the classical kinetic energy formula as K classical = 1 2 mv 2 = 1 2 ( mc 2 ) µ v 2 c 2 = 1 2 ( mc 2 ) β 2 . (a) If K classical =10 . 0MeV, then β = r 2 K classical mc 2 = r 2(10 . 0MeV) 0 . 511 MeV =6 . 256 , which, of course, is impossible (see the Ultimate Speed subsection of § 38-2). If we use this value anyway, then the classical orbital radius formula yields r = mv | q | B = mβc eB = ( 9 . 11 × 10 31 kg ) (6 . 256) ( 2 . 998 × 10 8 m / s ) (1 . 6 × 10 19 C) (2 . 20 T) =4 . 85 × 10 3 m . If, however, we use the correct value for β (calculated in the next part) then the classical radius formula would give about 0 . 77 mm. (b) Before using the relativistically correct orbital radius formula, we must compute
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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